Integrand size = 29, antiderivative size = 271 \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=-\frac {(c-11 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-d)^{7/2} f}-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}-\frac {d (3 c+7 d) \cos (e+f x)}{6 a (c-d)^2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{6 a (c-d)^3 (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \] Output:
-1/4*(c-11*d)*arctanh(1/2*a^(1/2)*(c-d)^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin( f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))*2^(1/2)/a^(3/2)/(c-d)^(7/2)/f-1/2*co s(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(3/2)-1/6*d*(3*c+ 7*d)*cos(f*x+e)/a/(c-d)^2/(c+d)/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^ (3/2)-1/6*d*(3*c^2+38*c*d+19*d^2)*cos(f*x+e)/a/(c-d)^3/(c+d)^2/f/(a+a*sin( f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)
Time = 6.34 (sec) , antiderivative size = 478, normalized size of antiderivative = 1.76 \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 c^4+12 c^3 d+81 c^2 d^2+70 c d^3+11 d^4-d^2 \left (3 c^2+38 c d+19 d^2\right ) \cos (2 (e+f x))+12 d \left (c^3+8 c^2 d+9 c d^2+2 d^3\right ) \sin (e+f x)\right )}{(c+d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}+\frac {3 (c-11 d) \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}\right )}{12 (c-d)^3 f (a (1+\sin (e+f x)))^{3/2} \sqrt {c+d \sin (e+f x)}} \] Input:
Integrate[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^(5/2)),x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*(-(((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*c^4 + 12*c^3*d + 81*c^2*d^2 + 70*c*d^3 + 11*d^4 - d^2*(3*c^2 + 38*c*d + 19*d^2)*Cos[2*(e + f*x)] + 12*d*(c^3 + 8*c^2*d + 9*c*d^2 + 2*d^3 )*Sin[e + f*x]))/((c + d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c + d*S in[e + f*x]))) + (3*(c - 11*d)*(Log[1 + Tan[(e + f*x)/2]] - Log[c - d + 2* Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]]))/(Sec[(e + f*x)/2]^2/(2 + 2*Tan[(e + f*x)/2]) - (-1 /2*((c - d)*Sec[(e + f*x)/2]^2) + (Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^( 3/2)*(d + d*Cos[e + f*x] + c*Sin[e + f*x]))/Sqrt[c + d*Sin[e + f*x]])/(c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]))))/(12*(c - d)^3*f*(a*(1 + Sin[e + f*x]))^(3/ 2)*Sqrt[c + d*Sin[e + f*x]])
Time = 1.39 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3245, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3261, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle -\frac {\int -\frac {a (c-7 d)+4 a d \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{5/2}}dx}{2 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (c-7 d)+4 a d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{5/2}}dx}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (c-7 d)+4 a d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{5/2}}dx}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {\left (3 c^2-24 d c-19 d^2\right ) a^2+2 d (3 c+7 d) \sin (e+f x) a^2}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\left (3 c^2-24 d c-19 d^2\right ) a^2+2 d (3 c+7 d) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (3 c^2-24 d c-19 d^2\right ) a^2+2 d (3 c+7 d) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {-\frac {2 \int -\frac {3 a^3 (c-11 d) (c+d)^2}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a \left (c^2-d^2\right )}-\frac {2 a^2 d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 a^2 (c-11 d) (c+d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}-\frac {2 a^2 d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 a^2 (c-11 d) (c+d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}-\frac {2 a^2 d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {-\frac {6 a^3 (c-11 d) (c+d)^2 \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f \left (c^2-d^2\right )}-\frac {2 a^2 d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {-\frac {3 \sqrt {2} a^{3/2} (c-11 d) (c+d)^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f \sqrt {c-d} \left (c^2-d^2\right )}-\frac {2 a^2 d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{3 a \left (c^2-d^2\right )}-\frac {2 a d (3 c+7 d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}\) |
Input:
Int[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^(5/2)),x]
Output:
-1/2*Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x ])^(3/2)) + ((-2*a*d*(3*c + 7*d)*Cos[e + f*x])/(3*(c^2 - d^2)*f*Sqrt[a + a *Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) + ((-3*Sqrt[2]*a^(3/2)*(c - 11* d)*(c + d)^2*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt[c - d]*(c^2 - d^2)*f) - (2*a^2*d*(3*c^2 + 38*c*d + 19*d^2)*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a *Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))/(3*a*(c^2 - d^2)))/(4*a^2*(c - d ))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Timed out.
\[\int \frac {1}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]
Input:
int(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x)
Output:
int(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 1474 vs. \(2 (236) = 472\).
Time = 0.65 (sec) , antiderivative size = 3182, normalized size of antiderivative = 11.74 \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fr icas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="ma xima")
Output:
integrate(1/((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^(5/2)), x)
\[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="gi ac")
Output:
integrate(1/((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^(5/2)), x)
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))^(5/2)),x)
Output:
int(1/((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))^(5/2)), x)
\[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{5} d^{3}+3 \sin \left (f x +e \right )^{4} c \,d^{2}+2 \sin \left (f x +e \right )^{4} d^{3}+3 \sin \left (f x +e \right )^{3} c^{2} d +6 \sin \left (f x +e \right )^{3} c \,d^{2}+\sin \left (f x +e \right )^{3} d^{3}+\sin \left (f x +e \right )^{2} c^{3}+6 \sin \left (f x +e \right )^{2} c^{2} d +3 \sin \left (f x +e \right )^{2} c \,d^{2}+2 \sin \left (f x +e \right ) c^{3}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x \right )}{a^{2}} \] Input:
int(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1))/(sin(e + f* x)**5*d**3 + 3*sin(e + f*x)**4*c*d**2 + 2*sin(e + f*x)**4*d**3 + 3*sin(e + f*x)**3*c**2*d + 6*sin(e + f*x)**3*c*d**2 + sin(e + f*x)**3*d**3 + sin(e + f*x)**2*c**3 + 6*sin(e + f*x)**2*c**2*d + 3*sin(e + f*x)**2*c*d**2 + 2*s in(e + f*x)*c**3 + 3*sin(e + f*x)*c**2*d + c**3),x))/a**2